Problem: $c(n) = -6 \left(-\dfrac{1}{3}\right)^{n - 1}$ What is the $2^\text{nd}$ term in the sequence?
Solution: This is an explicit formula. All we have to do is plug $n=2$ in the formula to find the $2^\text{nd}$ term. $\begin{aligned} c({2}) &=-6\left(-\dfrac13\right)^{{2} - 1} \\\\ &= 2 \end{aligned}$ The $2^\text{nd}$ term is $2$.